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The henry's law constant for the solubil...

The henry's law constant for the solubility of `N_2` gas in water at 298 K is `1.0xx10^5` atm . The mole fraction of `N_2 ` in air is 0.8 . The number of moles of `N_2` from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

A

`4.0 xx10^(-4)`

B

`4.0 xx10 ^(-5)`

C

`5.0 xx 10^(-4)`

D

`4.0 xx 10 ^(-6)`

Text Solution

Verified by Experts

As per Henry’s law :
`pN_2 =K_H(x N_2) ` in solution
Now, `pN_2 = p _("Total") xx (xN_2)` In air `= 5xx 0.8 = 4 atm`
`rArr 4 = 10^5 xx(xN_2)` in solution
`rArr (xN_2)` in solution `=4xx10(-5) = (""^nN_2)/(""^nN_2+""^nO_2) ~~(""^nN_2)/(""^nO_2)`
`rArr (xN_2)` in solution `=4xx10^(-5) xx ""^nO_2 = 4xx 10^(-4)`
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